pgm to find middle element of linklist(in efficent manner)
Answer Posted / ashwini
struct node
{
int data;
struct node *ptr;
};
struct node mid_element(struct node* head)//since we pass addr
{
int count=0,n_count,i=0;
struct node* temp,*mid;
temp=mid=head;
while(temp -> ptr != NULL)
{
count++;
temp = temp->otr;
}
count++;
if(count % 2)
{
n_count = (count/2)+1;
for(i=0 ; i<n_count ; i++)
mid = mid -> ptr;
}
return mid;
}
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In cryptography, you could often break the algorithm if you know what was the original (plain) text that was encoded into the current ciphertext. This is called the plain text attack. In this simple problem, we illustrate the plain text attack on a simple substitution cipher encryption, where you know each letter has been substituted with a different letter from the alphabet but you don’t know what that letter is. You are given the cipherText as the input string to the function getwordSets(). You know that a plain text "AMMUNITION" occurs somewhere in this cipher text. Now, you have to find out which sets of characters corresponds to the encrypted form of the "AMMUNITION". You can assume that the encryption follows simple substitution only. [Hint: You could use the pattern in the "AMMUNITION" like MM occurring twice together to identify this]
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